Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider a continuous function $f:\mathbb{R}\rightarrow[0,\infty)$ that does not tend to zero as its argument tends to infinity. Formally, there is some $\varepsilon>0$ such that there does not exist a $T\in\mathbb{R}$ for which

$$t\geq T\Rightarrow f(t)\leq \varepsilon.$$

Is it true that there exists some $\alpha>0$ such that for any $t_0,t_1\in\mathbb{R}$


I've been trying to think of a counterexample, but continuity is making it difficult.


EDIT: Sorry, the question was meant to be:

"Is it true that there exists some $\alpha>0$ such that for any $t_0,t_1\in\mathbb{R}$, with $t_1\geq t_0$, one can always find a sufficiently large $t_2\geq t_1$ so that


share|cite|improve this question
Drawing a picture of what's going on may help here. – tharris Apr 22 '13 at 15:48
up vote 1 down vote accepted

You can just make the integral zero for some choices of $t_i$ by making $f$ itself zero at such places. Then you only need to add lots of spikes to make it not converge to 0.

Edit: enter image description here

Second edit for edited question: Just place the spikes so that they are identical, area 1, and centred at $x=n^2$ for each $n\in\mathbb Z$. For bad choices of $t_0$, integral is effectively bounded by something like $t_1^{1/2}-t_0^{1/2}$ so the statement is whether there is $\alpha>0$ such that you can find $t_1\ge t_0$ $t_1^{1/2}-t_0^{1/2} \ge \alpha (t_1-t_0) \iff \alpha^{-1} \ge t_1^{1/2} + t_0^{1/2}$ regardless of how big $t_0$ is. This clearly does not hold.

share|cite|improve this answer
Agreed, however I'm having trouble adding the spikes without breaking continuity. If you had a fixed $\alpha$ this would be easy, but its the "for any" that is causing me trouble. Can think of an actual counterexample? – jkn Apr 22 '13 at 15:57
@jkn try $f_n$ with support in $[n,n+1/n^\beta]$ for a suitably chosen $\beta \geq 1$, and then define $f = \sum_n f_n$ (which happens to equal $\sup_n f_n$ which may or may not be useful to you). – kahen Apr 22 '13 at 16:09
@Sharkos, aha! Thanks. – jkn Apr 22 '13 at 16:21
@jkn - spikes as shown are obviously continuous. If you make $f([a,b]) = 0$ then $\int_{t_0}^{t_1} f = 0$ for all $t_i\in [a,b]$ so you don't care about $\alpha$. – Sharkos Apr 22 '13 at 16:21
I'm not revising this answer any more times, this is getting silly... – Sharkos Apr 22 '13 at 16:45

It is false. If it were true then $$ \lim_{t\to\infty}\int_{t_0}^tf(s)\,ds=\infty, $$ which is not the case if $f$ is integrable (in the Lebesgue sense, or as an improper integral if you are talking about Riemann integrals) on $\mathbb{R}$.


Let $$ T(x)=\begin{cases} 1-|x| & \text{if }|x|\le1,\\0 & \text{if }|x|>1, \end{cases} $$ and define $$ f(x)=\sum_{n=1}^\infty T(2^n(x-n)). $$ $T(2^n(x-n))$ is supported on $[n-2^{-n},n+2^{-n}]$ and its integral is $2^{-n}$, so that $$ \int_{-\infty}^{+\infty}f(x)\,dx=\sum_{n=1}^\infty2^{-n}=1\text{ and }f(n)=1. $$

share|cite|improve this answer
Can't the unsigned Lebesgue integral take values in $[0,+\infty]$? What do you mean by integrable here? – jkn Apr 22 '13 at 16:00
Presumably that it is in $\mathcal L^1$? – kahen Apr 22 '13 at 16:04
(I'm not assuming $f\in\mathcal{L}^1$, all I'm assuming is that $f\in\mathcal{C}$). – jkn Apr 22 '13 at 16:09
If Lebesgue integral: $f$ is measurable and $\int_{-\infty}^{+\infty}f(x)\,dx<\infty$. If Riemann integral: $\int_{-\infty}^{+\infty}f(x)\,dx$ exists as an improper integral- – Julián Aguirre Apr 22 '13 at 16:12
Pretty sure the examples you've given tend to 0 as $x\to\infty$. – Sharkos Apr 22 '13 at 16:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.